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\input{wang_preamble.tex}

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%\usepackage{xeCJK} % 处理中文、日文和韩文（统称为 CJK 文字）的排版
%\setCJKmainfont{SimSun} % 设置正文字体为宋体
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\usepackage{xeCJK} % 支持中文字体
\setCJKmainfont{Songti SC} % 设置主要中文字体，用于正文中的中文文本。设置主要中文字体为宋体
%\setCJKmonofont{Menlo} % 设置等宽中文字体，用于代码块、等宽文本等。设置等宽中文字体为 Menlo
%\setCJKsansfont{PingFang SC} % 设置无衬线中文字体，用于标题、图表标签等。设置无衬线中文字体为 PingFang SC
%\setCJKromanfont{Songti SC} % 设置罗马中文字体，用于罗马字体中的中文文本。设置罗马中文字体为宋体

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%\newcommand{\showsolution}{0} %%设置showsolution=0, 编译生成试卷
\newcommand{\showsolution}{1} %%设置showsolution=1, 编译生成试卷解答、考完发给学生
%\newcommand{\showsolution}{2} %%设置showsolution=2, 编译生成试卷解答与评分标准、阅卷与试卷袋归档

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%% 填写课程信息：
\newcommand{\CourseName}{高等代数复习2}
\newcommand{\CourseStudents}{2024 级数学与应用数学1班}
\newcommand{\ExamContents}{本次数学考试的主要内容是：
}

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\usepackage{titling}
\setlength{\droptitle}{-2cm}   % 标题上移2cm

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\begin{document}

\renewcommand{\today}{\number\year \,年 \number\month \,月 \number\day \,日}
\date{2024年12月24日} %考试日期
\thispagestyle{fancy} % 第一页也显示“第几页，共几页”的信息。

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%% 试卷：
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\author{学号 \underline{\hspace{3.5cm}} 姓名 \underline{\hspace{3.5cm}} }
\title{\CourseName }
\maketitle

\fi

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%%试卷解答：
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\author{\CourseStudents}
\title{\CourseName 解答 }
\maketitle

%\abstract{\ExamContents }

\fi

\thispagestyle{fancy} % 第一页也显示“第几页，共几页”的信息。

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%选择题开始
\begin{enumerate}

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%\newpage 
\item  %选择题第1题
设 $A, B, C$ 均为 $n$ 阶矩阵，则下列命题正确的是
%
\ifnum\showsolution=0
\dotfill (\hspace{1cm})
\fi
%
\begin{tasks}(1) %每行1个选项
\task [A.]  若 $A$ 是可逆矩阵，则从 $AB=AC$ 可推出 $BA=CA$.
\task [B.]  若 $A$ 是可逆矩阵，则 $AB=BA$.
\task [C.]  若 $A \neq O$，则从 $AB=AC$ 可推出 $B=C$.
\task [D.]  若 $B\neq C$，则 $AB\neq AC$.
\end{tasks}

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%%选择题第1题：解答与评分标准
\ifnum\showsolution>0

{\color{red} 解答：A. 
\begin{tasks}(1) %每行1个选项
\task [A.]  若 $A$ 是可逆矩阵，若 $AB=AC$，则 $A^{-1}AB=A^{-1}AC$, 即 $B=C$. 于是 $BA=CA$.  
\task [B.]  可以找到例子，$A$ 可逆，但 $AB\neq BA$. 
\task [C.]  可以找到反例， $A \neq O$，$B\neq C$, 但 $AB=AC$. 
\task [D.]  可以找到反例， $B\neq C$，但 $AB=AC$. 
\end{tasks}

}

\fi

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%\newpage 
\item  %选择题第2题
设 $A$ 为 $n$ 阶方阵，$A$ 适合下列哪个条件时，$E_n-A$ 必是可逆矩阵。
%
\ifnum\showsolution=0
\dotfill (\hspace{1cm})
\fi
%
\begin{tasks}(2) %每行2个选项
\task [A.]  $A^n=O$ 
\task [B.]  $A$ 可逆
\task [C.]  $|A|=0$
\task [D.]  $A$ 的主对角线元素为 0
\end{tasks}

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%%选择题第2题：解答与评分标准
\ifnum\showsolution>0

{\color{red} 解答：A. 
\begin{tasks}(1) %每行1个选项
\task [A.]  若 $A^n=O$, 则 $(E-A)(E+A+A^2+\cdots+A^{n-1})=E-A^n=E$. 因此 $E-A$ 可逆。
\task [B.]  例如 $A=E$ 可逆，但此时 $E-A=O$ 不可逆。
\task [C.]  可以找到例子，$|A|=0$ 但 $E-A$ 不可逆。
\task [D.]  可以找到例子，$A$ 的主对角线元素为 0, 但 $E-A$ 不可逆。
\end{tasks}
}

\fi

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%\newpage 
\item  %选择题第3题

设 $A$ 为 $n$ 阶方阵，则下列结论错误的是
%
\ifnum\showsolution=0
\dotfill (\hspace{1cm})
\fi
%
\begin{tasks}(2) %每行2个选项
\task [A.] 若 $|A|\neq 0$，则 $A^*$ 可逆。
\task [B.] 若 $|A|= 0$，则 $A^*$ 不可逆。
\task [C.] 若 $|A^*|\neq 0$，则 $A$ 可逆。
\task [D.] $|AA^*|=|A|$. 
\end{tasks}

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%%选择题第3题：解答与评分标准
\ifnum\showsolution>0

{\color{red} 解答：D. 
\begin{tasks}(1) %每行1个选项
\task [A.]  若 $d=|A|\neq 0$，则从 $AA^*=dE$ 可得 $(d^{-1}A)A^*=E$, 因此 $A^*$ 的逆阵是 $d^{-1}A$. 
\task [B.]  若 $d=|A|= 0$，则 $AA^*=dE=O$; 若 $A^*$ 可逆，则 $A=O$; 但这时 $A^*=O$. 
\task [C.]  反证：若 $A$ 不可逆，则 $d=|A|=0$, 此时 $AA^*=dE=O$. 根据条件 $|A^*|\neq 0$ 可得 $A^*$ 可逆。则从 $AA^*=O$ 可得 $A=O$, 但这时 $A^*=O$, 这与条件 $|A^*|\neq 0$ 矛盾。
\task [D.]  记 $d=|A|$, 则 $|AA^*|=|dE|=d^n$. 注意 $dE$ 是对角线元素为 $d$ 的数量矩阵。
\end{tasks}
}

\fi
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%\newpage 
\item  %选择题第4题
设 $A, B$ 均为 $n$ 阶方阵，则下述哪个一定成立？
%
\ifnum\showsolution=0
\dotfill (\hspace{1cm})
\fi
%
\begin{tasks}(2) %每行2个选项
\task [A.]  $R(A+B)\geq R(A)+R(B)$.
\task [B.]  $R(A+B)\leq R(A)+R(B)$.
\task [C.]  $R(A+B)\geq \max\{R(A)+R(B)\}$.
\task [D.]  $R(A+B)< \min\{R(A)+R(B)\}$.
\end{tasks}

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%%选择题第4题：解答与评分标准
\ifnum\showsolution>0

{\color{red} 解答：B. 
记矩阵 $A$ 的行向量组为 $\alpha_1, \alpha_2, \cdots, \alpha_n$, 一个极大线性无关组为 $\alpha_{i_1}, \cdots, \alpha_{i_s}$, 此时 $R(A)=s$. 记矩阵 $B$ 的行向量组为 $\beta_1,\beta_2, \cdots, \beta_n$, 一个极大线性无关组为 $\beta_{j_1}, \cdots, \beta_{j_t}$, 此时 $R(B)=t$. 
则矩阵 $A+B$ 的行向量组为 $\alpha_1+\beta_1, \alpha_2+\beta_2, \cdots, \alpha_n+\beta_n$. 
且每个向量 $\alpha_k+\beta_k$ 都可以由 $\alpha_{i_1}, \cdots, \alpha_{i_s}, \beta_{j_1}, \cdots, \beta_{j_t}$ 线性表示。
因此矩阵 $A+B$ 的行向量组的秩小于等于 $s+t$. 
}

\fi

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%\newpage 
\item  %选择题第5题
求下述齐次线性方程组的基础解系：
%
\ifnum\showsolution=0
\dotfill (\hspace{1cm})
\fi
%
$$\left\{
  \begin{array}{c}
    x_1-x_2=0, \\
    x_2-x_3=0, \\
    x_1-x_3=0. \\
  \end{array}
\right.
$$ 

\begin{tasks}(2) %每行2个选项
\task [A.]  $(2,2,2)^T, (1,1,0)^T$
\task [B.]  $(1,1,1)^T$
\task [C.]  $(4,4,4)^T, (0,1,1)^T$
\task [D.]  $(1,0,1)^T$
\end{tasks}
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%%选择题第5题：解答与评分标准
\ifnum\showsolution>0

{\color{red} 解答：B. 
通解为 $(x_1,x_2,x_3)^T=(k,k,k)$, 其中 $k$ 为任意实数。因此一个基础解系为 $(1,1,1)^T$. 
}

\fi

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%\newpage 
\item  %选择题第6题
设下述线性方程组无解，求 $k$ 的值：
\ifnum\showsolution=0 \dotfill (\hspace{1cm}) \fi

$$\left\{
  \begin{array}{rcl}
    x_1+2x_2-x_3 &=&4,\\
    x_2+2x_3&=&2, \\
    (k-1)(k-2)x_3&=&(k-1)(k-4). \\
  \end{array}
\right.
$$
 
\begin{tasks}(4) %每行2个选项
\task [A.]  $2$
\task [B.]  $3$
\task [C.]  $4$
\task [D.]  $1$
\end{tasks}
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%%选择题第6题：解答与评分标准
\ifnum\showsolution>0

{\color{red} 解答：A. 
将增广矩阵 $(A,\beta)$ 化为行阶梯形。可见当 $k=1$ 时 $R(A,\beta)=R(A)=2$, 此时有无穷多解；
当 $k=2$ 时 $R(A)=2$ 但 $R(A,\beta)=3$, 此时无解；
当 $k\neq 1,2$ 时，$R(A,\beta)=R(A)=3$, 此时有唯一解。
}

\fi

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%\newpage 
\item  %选择题第7题
设 $A$ 为 $m\times n$ 矩阵，$R(A)=m<n$，则下列结论正确的是
%
\ifnum\showsolution=0
\dotfill (\hspace{1cm})
\fi
%
\begin{tasks}(2) %每行2个选项
\task [A.]  $Ax=0$ 只有零解。
\task [B.]  非齐次线性方程组 $Ax=b$ 有无穷多解。
\task [C.]  $A$ 中任一个 $m$ 阶子式均不等于零。
\task [D.]  $A$ 中任意 $m$ 个列向量线性无关。
\end{tasks}
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%%选择题第7题：解答与评分标准
\ifnum\showsolution>0

{\color{red} 解答：B. 
\begin{tasks}(1) %每行1个选项
\task [A.]  齐次线性方程组 $Ax=0$ 的未知数个数为 $n$. 因为 $R(A)=m<n$, 所以有无穷多解。 
\task [B.]  因为增广矩阵 $(A,b)$ 也只有 $m$ 行，从 $R(A)=m$ 可知时 $R(A,b)=R(A)=m$. 因为 $m<n$ 所以 $Ax=b$ 有无穷多解。
\task [C.]  按照秩的定义，$R(A)=m$ 是指 $A$ 中存在一个 $m$ 阶子式，其值不等于零。
\task [D.]  矩阵 $A$ 的列数为 $n$ 而 $n>m$, 所以可以有某 $m$ 个列向量线性无关，而另外 $m$ 个列向量线性相关。
\end{tasks}
}

\fi


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%\newpage 
\item  %选择题第8题
设向量组 $\alpha_1, \alpha_2, \alpha_3$ 是齐次线性方程组 $Ax=0$ 的一个基础解系，则下列哪个也是 $Ax=0$ 的基础解系？
%
\ifnum\showsolution=0
\dotfill (\hspace{1cm})
\fi
%
\begin{tasks}(2) %每行2个选项
\task [A.]  $\alpha_1+\alpha_2, \alpha_2+\alpha_3, \alpha_3-\alpha_1$
\task [B.]  $\alpha_1+\alpha_2, \alpha_2+\alpha_3, \alpha_1+2\alpha_2+\alpha_3$
\task [C.]  $2\alpha_1, \alpha_1+\alpha_2, \alpha_1-\alpha_2$
\task [D.]  $\alpha_1+\alpha_2, \alpha_1-\alpha_2, \alpha_3$
\end{tasks}
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%%选择题第8题：解答与评分标准
\ifnum\showsolution>0

{\color{red} 解答：D. 
\begin{tasks}(1) %每行1个选项
\task [A.]  向量组 $\alpha_1+\alpha_2, \alpha_2+\alpha_3, \alpha_3-\alpha_1$ 是线性相关的。
\task [B.]  向量组 $\alpha_1+\alpha_2, \alpha_2+\alpha_3, \alpha_1+2\alpha_2+\alpha_3$ 是线性相关的。
\task [C.]  向量组 $2\alpha_1, \alpha_1+\alpha_2, \alpha_1-\alpha_2$ 是线性相关的。
\task [D.]  向量组 $\alpha_1+\alpha_2, \alpha_1-\alpha_2, \alpha_3$ 与基础解系 $\alpha_1, \alpha_2, \alpha_3$ 可以相互线性表示，而且是线性无关的，因此这个向量组也是基础解系。
\end{tasks}

}

\fi


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%\newpage 
\item  %选择题第9题
设向量 $\xi_1, \xi_2$ 是非齐次线性方程组 $Ax=b$ 的两个不同的解，$\eta$ 是其导出组 $Ax=0$ 的一个非零解,则 
%
\ifnum\showsolution=0
\dotfill (\hspace{1cm})
\fi
%
\begin{tasks}(1) %每行1个选项
\task [A.]  $\xi_1-\xi_2, \xi_1$ 线性无关
\task [B.]  $\xi_1-\xi_2, \eta$ 线性相关
\task [C.]  $Ax=b$ 的通解为 $\xi_1+k\eta$, $k$ 为任意常数
\task [D.]  $Ax=b$ 的通解为 $\xi_1+s(\xi_1-\xi_2)+t\eta$, $s, t$ 为任意常数
\end{tasks}
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%%选择题第9题：解答与评分标准
\ifnum\showsolution>0

{\color{red} 解答：A. 
\begin{tasks}(1) %每行1个选项
\task [A.]  反证：若 $\xi_1-\xi_2, \xi_1$ 线性相关。设 $\xi_1-\xi_2=k\xi_1$, 则左乘 $A$ 可得 $A(\xi_1-\xi_2)=A(k\xi_1)$, 即得 $b-b=kb$. 因为题目说 $Ax=b$ 是非齐次线性方程组，所以 $b\neq 0$. 所以 $k=0$. 这样就有 $\xi_1=\xi_2$, 与题目矛盾。另一种情况，设 $\xi_1=k(\xi_1-\xi_2)$, 同样左乘 $A$ 可得 $A\xi_1= A[k(\xi_1-\xi_2)]$, 得到 $b=0$, 也与题目矛盾。
\task [B.]  因为 $A(\xi_1-\xi_2)=A\xi_1-A\xi_2=b-b=0$, 所以 $\xi_1-\xi_2$ 是 $Ax=0$ 的一个非零解。$\eta$ 也是 $Ax=0$ 的一个非零解。向量组 $\xi_1-\xi_2, \eta$ 线性相关和线性无关都可能。
\task [C.]  向量 $\xi_1$ 是 $Ax=b$ 的一个特解，但导出组 $Ax=0$ 的基础解系可能包含不止一个向量。
\task [D.]  向量 $\xi_1-\xi_2, \eta$ 都是导出组 $Ax=0$ 的非零解，但是这两个向量可能线性相关。即使这两个向量线性无关，导出组 $Ax=0$ 的基础解系也可能包含不止两个向量。
\end{tasks}

}

\fi

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%\newpage 
\item  %选择题第10题
若向量 $\beta$ 可由向量组 $\alpha_1, \cdots, \alpha_s$ 线性表示，则
%
\ifnum\showsolution=0
\dotfill (\hspace{1cm})
\fi
%
\begin{tasks}(1) %每行2个选项
\task [A.]  存在一组不全为零的数 $k_i \ (1\leq i\leq s)$ 使得 $\beta=\sum\limits_{i=1}^s k_i\alpha_i$. 
\task [B.]  对 $\beta$ 线性表示式不唯一。
\task [C.]  $\beta, \alpha_1, \cdots, \alpha_s$ 线性相关。
\task [D.]  存在一组全为零的数 $k_i \ (1\leq i\leq s)$ 使得 $\beta=\sum\limits_{i=1}^s k_i\alpha_i$. 
\end{tasks}
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%%选择题第10题：解答与评分标准
\ifnum\showsolution>0

{\color{red} 解答：C. 
\begin{tasks}(1) %每行2个选项
\task [A.]  若 $\beta$ 是零向量，且向量组 $\alpha_1, \cdots, \alpha_s$ 线性无关，则使得 $\beta=\sum\limits_{i=1}^s k_i\alpha_i$ 的数 $k_i$ 只能全为零。
\task [B.]  若向量组 $\alpha_1, \cdots, \alpha_s$ 线性无关，则用这组向量来表示 $\beta$ 的方式是唯一的。
\task [C.]  因为 $\beta-k_1\alpha_1-\cdots-k_s\alpha_s=0$,且系数不全为零，所以向量组 $\beta, \alpha_1, \cdots, \alpha_s$ 线性相关。
\task [D.]  这时 $\beta$ 只能为零向量，但题目中没有这个条件。另外，数字零总是存在的，不需要另加条件说明零的存在性。
\end{tasks}
}

\fi

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%\newpage 
\item  %选择题第11题
设 $A, B$ 分别是 $m\times n$ 和 $n\times m$ 矩阵，则
%
\ifnum\showsolution=0
\dotfill (\hspace{1cm})
\fi
%
\begin{tasks}(1) %每行2个选项
\task [A.]  当 $m>n$ 时，$|AB|\neq 0$.
\task [B.]  当 $m>n$ 时，$|AB|= 0$.
\task [C.]  当 $m<n$ 时，$|AB|\neq 0$.
\task [D.]  当 $m<n$ 时，$|AB|= 0$.
\end{tasks}
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%%选择题第11题：解答与评分标准
\ifnum\showsolution>0

{\color{red} 解答：B. 
\begin{enumerate}[label=(\arabic*)]
\item  乘积 $AB$ 是 $m\times m$ 矩阵，当 $m>n$ 时，因为 $R(AB)\le R(A)\le n<m$, 所以有 $|AB|=0$. 
\item  当 $m<n$ 时，$R(AB)=m$ 与 $R(AB)<m$ 都有可能，因此 $|AB|\neq 0$ 与 $|AB|=0$ 都有可能。
\end{enumerate}

}

\fi

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%选择题结束
\end{enumerate}

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\end{document}





